Problem:
Find ax5+by5 if the real numbers a,b,x and y satisfy the equations
ax+by=3,ax2+by2=7,ax3+by3=16,ax4+by4=42
Solution:
For n=1 and n=2, the identity
(axn+1+byn+1)(x+y)−(axn+byn)xy=axn+2+byn+2(1)
yields the equations
7(x+y)−3xy=16 and 16(x+y)−7xy=42
Solving these last two equations simultaneously, one finds that
x+y=−14 and xy=−38(2)
Applying (1) with n=3 then gives
ax5+by5=(42)(−14)−(16)(−38)=−588+608=20
Note: From (2) we can solve for x and y. We obtain x=−7±87 and y=−7∓87, from which a=7649±661245787 and b=7649∓661245787.
The problems on this page are the property of the MAA's American Mathematics Competitions